| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 12 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 Part (i) is a standard trajectory derivation requiring substitution of parametric equations. Part (ii) involves treating the trajectory as a quadratic in tan α and applying the discriminant condition, which is a moderately sophisticated algebraic manipulation. Part (iii) requires geometric insight to recognize that maximum range occurs at the envelope boundary, then solve a straightforward intersection problem. This is above-average difficulty due to the multi-step reasoning and the non-standard approach in part (ii), but the techniques are well within A-level mechanics scope. |
| Spec | 1.02q Use intersection points: of graphs to solve equations3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow R_{\text{down}} = \frac{ | -40\sin\beta - 40 | }{\cos^2\beta} = \frac{40(1+\sin\beta)}{1-\sin^2\beta} = \frac{40}{1-\sin\beta}\) B1 |
From Question 12 in the mark scheme:
**(i)** $x = 20\cos at$, $y = 20\sin at - 5t^2$ **B1 B1**
$y = 20\sin\alpha \cdot \frac{x}{20\cos\alpha} - 5\left(\frac{x}{20\cos\alpha}\right)^2 = x\tan\alpha - \frac{x^2}{80}(1+\tan^2\alpha)$ AG **M1 A1**
**(ii)** $x^2\tan^2\alpha - 80x\tan\alpha + x^2 + 80y = 0$ (Can be implied by what follows.) **B1**
Real roots $\Rightarrow 6400x^2 - 4x^2(x^2 + 80y) \geq 0$ **M1 A1**
$\Rightarrow 1600 - x^2 - 80y \geq 0 \Rightarrow y \leq 20 - \frac{x^2}{80}$, $(x \neq 0)$ **A1**
**(iii)** $x = R\cos\beta$ and $y = R\sin\beta \Rightarrow y = x\tan\beta$
In $y = 20 - \frac{x^2}{80} \Rightarrow R\sin\beta = 20 - \frac{R^2(1-\sin^2\beta)}{80}$ **M1 A1**
$\therefore R^2(1-\sin^2\beta) + 80R\sin\beta - 1600 = 0 \Rightarrow (R[1-\sin\beta]+40)(R[1+\sin\beta]-40) = 0$ **M1**
$\Rightarrow R = \frac{40}{1+\sin\beta}$ (up) or $\frac{-40}{1-\sin\beta}$ (down) **A1**
**Alternative solution:**
$x^2 + 80\tan\beta\, x - 1600 = 0$ **B1**
$\Rightarrow x = -40\tan\beta \pm 40\sec\beta$ **B1**
$\Rightarrow R_{\text{up}} = \frac{-40\sin\beta + 40}{\cos^2\beta} = \frac{40(1-\sin\beta)}{1-\sin^2\beta} = \frac{40}{1+\sin\beta}$ **B1**
$\Rightarrow R_{\text{down}} = \frac{|-40\sin\beta - 40|}{\cos^2\beta} = \frac{40(1+\sin\beta)}{1-\sin^2\beta} = \frac{40}{1-\sin\beta}$ **B1**12 A particle is projected from the origin with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal.\\
(i) Prove that the equation of its trajectory is
$$y = x \tan \alpha - \frac { x ^ { 2 } } { 80 } \left( 1 + \tan ^ { 2 } \alpha \right) .$$
(ii) Regarding the equation of the trajectory as a quadratic equation in $\tan \alpha$, show that $\tan \alpha$ has real values provided that
$$y \leqslant 20 - \frac { x ^ { 2 } } { 80 } .$$
(iii) A plane is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The line $l$, with equation $y = x \tan 30 ^ { \circ }$, is a line of greatest slope in the plane. The particle is projected from the origin with speed $20 \mathrm {~ms} ^ { - 1 }$ from a point on the plane, in the vertical plane containing $l$. By considering the intersection of $l$ with the curve $y = 20 - \frac { x ^ { 2 } } { 80 }$, find the maximum range up this inclined plane.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q12 [12]}}